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प्रश्न
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficient if a coefficient can be repeated in an equation.
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उत्तर
Let the quadratic equation be
ax2 + bx + c = 0, a ≠ 0
| Coefficient | Values | Numbers of ways |
| a | 2, 4, 5 | 3 |
| b | 0, 2, 4, 5 | 4 |
| c | 0, 2, 4, 5 | 4 |
Required number = 3 × 4 × 4 = 48
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संबंधित प्रश्न
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Evaluate: 6!
Evaluate: 8! – 6!
Evaluate: (8 – 6)!
Compute: `(12/6)!`
Compute: `(9!)/(3! 6!)`
Compute: `(6! - 4!)/(4!)`
Compute: `(8!)/(6! - 4!)`
Write in terms of factorial:
6 × 7 × 8 × 9
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Find n, if `"n"/(8!) = 3/(6!) + 1/(4!)`
Find n, if `"n"/(6!) = 4/(8!) + 3/(6!)`
Find n, if (n + 1)! = 42 × (n – 1)!
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
Find n, if: `((15 - "n")!)/((13 - "n")!)` = 12
Show that: `(9!)/(3!6!) + (9!)/(4!5!) = (10!)/(4!6!)`
Find the value of: `(8! + 5(4!))/(4! - 12)`
Show that: `((2"n")!)/("n"!)` = 2n(2n – 1)(2n – 3)....5.3.1
